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Electronics and Communication Engineering (ECE) Exam  >  Electronics and Communication Engineering (ECE) Test  >  RRB JE Mock Test Series for ECE 2026  >  RRB JE ECE (CBT I) Mock Test- 6 - Electronics and Communication Engineering (ECE) MCQ

RRB JE ECE (CBT I) Mock Test- 6 Free Online Test 2026


Full Mock Test & Solutions: RRB JE ECE (CBT I) Mock Test- 6 (100 Questions)

You can boost your Electronics and Communication Engineering (ECE) 2026 exam preparation with this RRB JE ECE (CBT I) Mock Test- 6 (available with detailed solutions).. This mock test has been designed with the analysis of important topics, recent trends of the exam, and previous year questions of the last 3-years. All the questions have been designed to mirror the official pattern of Electronics and Communication Engineering (ECE) 2026 exam, helping you build speed, accuracy as per the actual exam.

Mock Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 90 minutes
  • - Total Questions: 100
  • - Analysis: Detailed Solutions & Performance Insights
  • - Sections covered: Mathematics

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RRB JE ECE (CBT I) Mock Test- 6 - Question 1
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Solve the following expression: 452 - 35 * 68 - 58 + 60% of 730.

Detailed Solution: Question 1

= 452 - 35 * 68 - 58 + 60% of 730

= 2025 - 2380 - 58 + 438

= 2463 - 2438

= 25

= 52

RRB JE ECE (CBT I) Mock Test- 6 - Question 2
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The elevation angles of the top of a tower 90 m high, from two points on the ground level on its opposite sides are 45° and 60°. What is the distance between the two points (rounded off to the nearest integer)? (√3 = 1.732)

Detailed Solution: Question 2

Let AB be the tower and C and D the two points.

Now AB/BC = tan⁡45∘=1

AB = BC = 90

Also AB/BD = tan⁡60∘ = √3

BD = 90/√3 = 30√3

Hence CD = CB + BD = 90 + 30√3 = 90 + 51.96 = 141.96 ≈ 142m

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RRB JE ECE (CBT I) Mock Test- 6 - Question 3
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Two quadratic equations: 16x2 - Ax + 25 and Bx2 + Ax + 4 = 0 have real and equal roots, then what is the value of √B?

Detailed Solution: Question 3

16x2 - Ax + 25 has real and equal roots

(-A)2 = 4 * 16 * 25

A = 40

Bx2 + Ax + 4 = 0 has real and equal roots

(A)2 = 4 * B * 4

(40)2 = 4 * B * 4

B = 100

√B = √100

√B = 10

RRB JE ECE (CBT I) Mock Test- 6 - Question 4
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Roshan deposited Rs 17500 in a bank that offers compound interest at 8% per annum. What is the interest obtained on the sum after 2 years?

Detailed Solution: Question 4

Amount after two years = 17500 x (1+8/100) 2

= 17500 x 1.08 x 1.08 = 20412

Interest = 20412 - 17500 = Rs 2912

RRB JE ECE (CBT I) Mock Test- 6 - Question 5
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Length of a rectangle is increased by 10% and breadth is decreased by 10%. Find the percent effect in the area.

Detailed Solution: Question 5

Let length = l

And breadth = b

Area = lb

After increment, length = l x (100 + 10)/100 = l x 110/100 = l x 11/10

After decrement, breadth = b x (100 - 10)/100 = b x 90/100 = b x 9/10

Area = l x 11/10 x b x 9/10

= 99lb/100 lb

% decrease = (lb - 99lb/100)/lb x 100

= (100lb - 99lb)/100 x 1/lb x 100

= lb/100 x 1/lb x 100

= 1%

RRB JE ECE (CBT I) Mock Test- 6 - Question 6
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One of the angles of a triangle is half the larger angle of a parallelogram. The respective ratio between the adjacent angles of the parallelogram is 14 : 31 . The smallest angle of the triangle is half the smaller angle of the parallelogram. What is the value of the largest angle of the triangle?

Detailed Solution: Question 6

We know that the sum of the adjacent angles of a parallelogram = 180o

Let the adjacent angles of the parallelogram be 14x and 31x

Therefore, 14x + 31x = 180o

45x = 180o

X = 4

RRB JE ECE (CBT I) Mock Test- 6 - Question 7
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Aman donated 10% of his Savings to Charity and gave 15% of his savings to his sister. Of the remaining savings he gave 60% to his son and gave 80% of the still remaining saving to his daughter. If the sum of the money given to the sister and the money left with Aman finally is Rs 84,000, what was his savings?

Detailed Solution: Question 7

Let his savings be Rs 100x

Money donated to charity = Rs 10x and Money given to sister = Rs 15x

Money left with Aman = 100x - 10x - 15x = Rs 75x

Money left with him after giving to son = (40/100)*75x = Rs 30x

Money left with him after giving to daughter = (20/100)*30x = Rs 6x

Given, 15x + 6x = 84000

=> 21x = 84000

=> x = 4000

His Savings = Rs 4,00,000

RRB JE ECE (CBT I) Mock Test- 6 - Question 8
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The ratio of the 4th and 2nd term of an A.P of positive terms is 5:3 and the product of the 1st and the 3rd terms is 72. What is the product of the 5th and 6th terms of the AP?

Detailed Solution: Question 8

Let the first term be 'a' and the common difference be 'd'.

Given, (a + 3d)/(a + d) = 5/3

=> 3a + 9d = 5a + 5d

=> 2a = 4d

=> a = 2d

Also, a(a + 2d) = 72

=> 2d(2d + 2d) = 72

=> 8d2 = 72

=> d2 = 9

=> d = 3

∴ a = 2*3 = 6

d ≠ −3 as the 1st term will become a = 2(−3) = − 6 but all terms are positive.

5th term = a + 4d = 6 + 4*3 = 18

6th term = 18 + 3 = 21

Required product = 18*21 = 378

RRB JE ECE (CBT I) Mock Test- 6 - Question 9
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Three pipes A, B and C can fill a cistern in 10 hrs. After working together for 4 hours, C is closed and A and B fill the cistern in 9 hrs. Then find the time in which the cistern can be filled by pipe C?

Detailed Solution: Question 9

A, B and C can fill in 1 hour = 1/10 of cistern

A, B and C can fill in 4 hour= 4/10 =2/5 of cistern

Remaining part = 1 - 2/5 =3/5 of cistern

3/5 of cistern is filled by A+B in 9 hours

Therefore, A and B can fill the cistern in 9*5/3 = 15 hrs

Hence in one hour C filled the cistern = 1/10 - 1/15 = 1/30 of cistern

Hence C can fill the cistern in 30 hrs.

RRB JE ECE (CBT I) Mock Test- 6 - Question 10
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If the average of twelve consecutive even numbers be 113, then find the sum of 3rd and 8th numbers among them?

Detailed Solution: Question 10

Let the 1st number = x

According to question,

x + (x+2) + (x+4) + ...... + (x+22) = 113*12

12x + 132 = 1356

12x = 1224

x = 102

Therefore 3rd number = 102 + 4 = 106

8th number = 102 + 14 = 116

Hence the required answer = 106 + 116 = 222

RRB JE ECE (CBT I) Mock Test- 6 - Question 11
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What approximate value should come in the place of the question mark in the following question ?

(72)2 ÷ ∛46656=?

Detailed Solution: Question 11

? = (72)2 ÷ ∛46656

? = (72)2 ÷ ∛(36x36x36) = (72)2 ÷ 36

? = (72 X 72)/36 = 144

RRB JE ECE (CBT I) Mock Test- 6 - Question 12
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Ranjan purchases an article at Rs.660 and marks up the price by 25% and after allowing the discount of Rs.99 he gains __% profit on the sale.

Given below are the steps involved. Arrange them in the sequential order.

A) Percent profit = (66/660) * 100 = 10%

B) Selling price of the item after discount of Rs.99 = 825 - 99 = Rs.726

C) Profit amount earned on the item = 726 - 660 = Rs.66

D) Cost price of the article is Rs.660 and marked price = 125% of 660 = Rs.825

Detailed Solution: Question 12

The correct arrangement is

D) Cost price of the article is Rs.660 and marked price = 125% of 660 = Rs.825

B) Selling price of the item after discount of Rs.99 = 825 - 99 = Rs.726

C) Profit amount earned on the item = 726 - 660 = Rs.66

A) Percent profit = (66/660) * 100 = 10%

RRB JE ECE (CBT I) Mock Test- 6 - Question 13
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Study the following table carefully and answer the related question.

The following table represents the three types of products A, B,.” and C manufactured in a company during Five months.

In which of the following month there is a maximum percentage increase/decrease in the manufacturing of all three types of products together?

Detailed Solution: Question 13

Total products manufactured

In January = 18 + 22 + 25 = 65

In February = 20 + 12 + 32 = 64

In March = 40 + 10 + 13 = 63

In April = 21 + 22 + 24 = 67

In May = 25 + 21 + 16 = 62

Percentage increase/decrease:

In February = ((65 - 64)/65) x 100 = 1.54%

In March = ((64 - 63)/64) x 100 = 1.56%

In April = ((67 - 63)/63) x 100 = 6.35%

In May = ((67 - 62)/67) x 100 = 7.46%

RRB JE ECE (CBT I) Mock Test- 6 - Question 14
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Two trains 140 m and 160 m long run at the speed of 60 kmph and 40 kmph respectively in opposite directions on parallel tracks. The time (in seconds) which they take to cross each other is :

Detailed Solution: Question 14

time = Distance/Velocity

Distance = 140+160 = 300m = 0.3km

Relative velocity = 60+40 = 100kmph (opp. direction) Time =0.3/100h

or 0.3/100 × 3600 = 10.8sec

Time =10.8 sec.

RRB JE ECE (CBT I) Mock Test- 6 - Question 15
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a2 + b2 = 5 and a3+b3 = 9. If ab = 2, what is the value of a4 + b4?

Detailed Solution: Question 15

Let (a+b) = c

c2 = (a+b)2 = a2+b2 + 2ab = 5 + 4 = 9

=> a + b = 3

Now, (a+b)(a3+b3) = a4+b4+ab(a2+b2)

=> 3 x 9 = a4+b4 + 2 x 5

=> a4+b4 = 17

RRB JE ECE (CBT I) Mock Test- 6 - Question 16
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The average weight of 24 students in a class is 40 kg. if the weight of the teacher is included, the average is increased by 500 gms. Find The weight of the teacher?

Detailed Solution: Question 16

25*40.5 - 24*40 = 52.5

Hence, the answer is option (2).

RRB JE ECE (CBT I) Mock Test- 6 - Question 17
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A shopkeeper makes a profit of 20% by selling a T.V. What would be the new profit percent if the shopkeeper paid 15% less for the T.V and the customer paid 10% more for the T.V?

Detailed Solution: Question 17

Let the price at which T.V was purchased by shopkeeper = Rs 100a

Hence the price at which T.V was sold by shopkeeper = Rs 120a

If shopkeeper paid 15% less than the previous amount hence new cost price of the T.V = Rs (100a X 0.85) = Rs 85a

New selling price of the T.V = Rs (120a X 1.1) = Rs 132a

Profit % = (132a-85a)/85a X 100 = 55.3%

RRB JE ECE (CBT I) Mock Test- 6 - Question 18
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Solve the following expression: (13/8) of (15/32) of 45% of 3072.

Detailed Solution: Question 18

= (13/8) of (15/32) of 45% of 3072

= (13/8) * (15/32) * (45/100) * 3072

= (13/8) * (3/32) * (9/4) * 307

= 13 * 3 * 9 * 3

= 1053

RRB JE ECE (CBT I) Mock Test- 6 - Question 19
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Following Pie Chart shows the percentage distribution of people of different ages in a Building. The table shows the ratio of Male and Female people in each age group. There is a total of 800 people in the building.

What is the difference between the total number of Males of age 20 to 60 and the total number of Females of age 20 to 60?

Detailed Solution: Question 19

Calculating the number of Male and Females of each age group

total number of Males of age 20 to 60 = 210 + 120 = 330

total number of Females of age 20 to 60 = 150 + 40 = 190

Required difference = 330 - 190 = 140

RRB JE ECE (CBT I) Mock Test- 6 - Question 20
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The population of a village is 6000. The next year, the number of male increases by 10% and the number of females increases by 20% and the population becomes 6840. What is the initial population of Male?

Detailed Solution: Question 20

Let the initial population of Males and Females be 100x and 100y respectively.

Given, 100x + 100y = 6000

=> x + y = 60..(i)

And, 110x + 120y = 6840

=> 11x + 12y = 684..(ii)

12*(i) - (ii), gives

x = 36

Initial population of Males = 3600

RRB JE ECE (CBT I) Mock Test- 6 - Question 21
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The average age of a Sheela and her sister is 16 years. The respective ratio of their ages is 7:1 respectively. Find Sheela's age?

Detailed Solution: Question 21

Sum of their ages = 2 X 16 = 32

Let 7x and x be their respective ages, then, 8x = 32 and x = 4

So, Sheela's age = 7x = 7 X 4 = 28 years . Hence, option 1.

RRB JE ECE (CBT I) Mock Test- 6 - Question 22
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20% of x% of 60 is equal to y% of 60% of 40, what is the value of the ratio x:y?

Detailed Solution: Question 22

20% of x% of 60 = y% of 60% of 40

x = 2y

x:y = 2:1

RRB JE ECE (CBT I) Mock Test- 6 - Question 23
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What is the remainder when x20 + x19 + x18 + x17 + x16 + x15 + x14 + x13 + x12 + x11 is divided by x9 + x8 + x7 + x6 + x5 + x4 + x3 + x2 + x + 1

Detailed Solution: Question 23

Numerator = x11 (x9 + x8 + x7 + x6 + x5 + x4 + x3 + x2 + x + 1) = x11 times the denominator.

The denominator divides the numerator, so the answer is 0.

RRB JE ECE (CBT I) Mock Test- 6 - Question 24
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A person invested equal amounts in two schemes A and B at the same rate of interest. Scheme A offers simple interest while scheme B offers compound interest. After two years he got Rs. 1920 from the Scheme A as interest and Rs. 2112 from scheme B. If the rate of interest is increased by 4%, what will be the total interest after two years of both schemes ?

Detailed Solution: Question 24

Simple Interest for 1 year = 1920/2 = Rs. 960

Compound Interest - Simple Interest = 2112 - 1920 = Rs. 912

Interest on Rs. 960 for 1 year = Rs. 912

= Rs. 4800

New rate = 24% per annum

S.I. = (4800×2×24)/100

= Rs. 2304

Compound Interest = 4800[(1+ 24/100)2 - 1)

= 4800[(1.24)2 - 1]4800

= Rs. 2580.48

Total interest = Rs. 4884.48

RRB JE ECE (CBT I) Mock Test- 6 - Question 25
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How many litres of a solution containing milk and water in the ratio 2:3 should be added to 30 litres of 70 % milk solution in order to get 60% milk solution?

Detailed Solution: Question 25

Let the number of liters of solution added be: x

The initial ratio of milk (m) to water (w) is:

=> m/w = 2/3

=> m = 2/5 of the solution

=> m = 40% of the solution

Acc. to the question, after adding x liters of 40% milk solution to 30 liters of 70% milk solution, we get 60% of (x+30) liters milk solution.

Hence, the equation becomes:

=> 40x + 30 * 70 = (x+30) * 60

=> 40x + 2100= 60x + 1800

=> 20x = 300

=> x = 15

RRB JE ECE (CBT I) Mock Test- 6 - Question 26
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If 12% of a is equal to 15% of b. Find the respective ratio of a and b.

Detailed Solution: Question 26

12% of a = 15% of b

=> a/b = 15/12

=> a/b = 5/4

=> a:b = 5:4

RRB JE ECE (CBT I) Mock Test- 6 - Question 27
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Downstream speed of boat A is 20% more than the downstream speed of boat B whose upstream speed is 6 km/hr less than upstream speed of boat A. What is the speed of boat A in still water? The stream for both boats is moving at 10 km/hr.

Detailed Solution: Question 27

Stream speed = 10 km/hr

Downstream speed of boat A = (6/5) x downstream speed of boat B

Upstream speed of boat A - upstream speed of boat B = downstream speed of boat A - downstream speed of boat B = 6

(6/5) x downstream speed of boat B - downstream speed of boat B = 6

Downstream speed of boat B = 30 km/hr

Downstream speed of boat A = 30 + 6 = 36 km/hr

Speed of boat A in still water = 36 - 10 = 26 km/hr

RRB JE ECE (CBT I) Mock Test- 6 - Question 28
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If the monthly income of a person is increased by 20% and his expenditures are increased by 40%, then his savings are decreased by what percent? (The ratio of income and expenditure is 4: 3 respectively.)

Detailed Solution: Question 28

Let income and expenditure are 'a' and 'b' respectively.

Savings = a - b = a - 3a/4 = a/4

When income and expenditure are increased by 20%. Then,

New income = 120% of a = 1.2a

New expenditure = 140% of b = 1.4 x 3a/4 = 2.1a/2

New savings = 1.2a - 2.1a/2 = 3a/20

Decrement in savings = (a/4) - (3a/20) = a/10

Percentage = a/10)/(a/4 x 100 = 40%

RRB JE ECE (CBT I) Mock Test- 6 - Question 29
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A and B can finish a work in 2 days, B and C in 3 days and A and C in 4 days. Find the ratio of time taken by A alone and C alone to finish the work?

Detailed Solution: Question 29

Suppose A, B and C take A, B and C days respectively to finish the work on their own.

1/A + 1/B = 1/2 (1)

1/B + 1/C = 1/3 (2)

1/A + 1/C = 1/4 (3)

(1)+(2)+(3) => 2(1/A+1/B+1/C) = 13/12

=> 1/A + 1/B + 1/C = 13/24 (4)

(4) - (2) => 1/A = 5/24

Thus A takes 24/5 = 4.8 days.

(4) - (1) => 1/C = 1/24

Hence C takes 24 days.

Required ratio = 4.8: 24 = 1:5

RRB JE ECE (CBT I) Mock Test- 6 - Question 30
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Which of the following is irrational?

Detailed Solution: Question 30

Product of (3 - √27) and (√9 + 3√3)

= (3 - 3√3)(3 + 3√3)

= - 18 (which is not irrational)

Option (2):

= Sum of (5 - 4√2) and 4(6 + √2)

= (5 - 4√2) + 4(6 + √2)

= 5 - 4√2 + 24 + 4√2

= 29 (which is not irrational)

Option (3):

= Sum of √4761 and √5175

= 69 + 15√23 (which is irrational)

Option (4):

= Product of √16875 and √27

= 75√3 x 3√3

= 675 (which is not irrational)

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About this Electronics and Communication Engineering (ECE) Mock Test

This mock test contains 100 questions designed to replicate the pattern, difficulty level and time pressure of the actual Electronics and Communication Engineering (ECE) exam. It is part of RRB JE Mock Test Series for ECE 2026 on EduRev, covering all the topics that are being asked as per the current Electronics and Communication Engineering (ECE) syllabus. Attempting this test under timed conditions will help you assess your exam readiness, identify subject-wise weak areas, and improve your speed and accuracy before the actual exam.

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If you found this RRB JE ECE (CBT I) Mock Test- 6 helpful, you can further enhance your Electronics and Communication Engineering (ECE) preparation with complete mock tests, previous year questions & practice tests available in RRB JE Mock Test Series for ECE 2026 course on EduRev. The course offers:
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